求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
问题描述:
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
答
tan(2π-a)cos(3π/2 -)cos(6π -a) / sin(+3π/2)cos(+3π/2) =(-tan)(-cosa)cosa/(-cosa)cosa=-tana